# Invariant subspaces for matrices via fixed points on Grassmannians

Let $$A$$ be an $$n \times n$$ invertible complex matrix. Let $$Gr(k)=Gr(k,\mathbb{C}^n)$$ be the complex $$k$$-Grassmannian, $$1\leq k \leq n$$. Since $$A$$ is invertible, it maps a $$k$$-dimensional subspace to a $$k$$-dimensional subspace, so it gives a function (which I'll call $$A_k$$) on $$Gr(k)$$. The fact that matrices have eigenvalues lets us deduce that $$A$$ has a $$k$$-dimensional invariant subspace - i.e. this map $$A_k$$ has a fixed point.

My question is this: is there a fixed point theorem on $$Gr(k)$$ (which ideally does not in any way depend on the existence of eigenvalues) that we can invoke to arrive at the same conclusion?

• This is slightly broken as written -- you need $M$ to be homotopic to the identity. Otherwise, consider the $2$-sphere (Euler characteristic $2$) and the antipodal map. But you are fine in this case, since $GL_n(\mathbb{C})$ is connected, $\bigwedge^k M$ is homotopic to $\bigwedge^k \text{Id}$. Feb 3 at 15:37